Integral of polynomial exponential function $\int_0^\infty e^{-ax}e^{c x^b } dx$

Question

Suppose $a,c >0$ and $b\in(0,1)$, then $$\int_0^\infty e^{-ax}e^{c x^b } dx ~?$$

My try:

I tried using integration by parts, but ending up with $e^{c x^b } $ again to be integrated again and again. Any other suggestions to get a closed-form expression for this integral? If not viable, then are there any simple upper bounds for this integral?

Thanks in advance for any help!

Answer 1

Hint: Expansion of $\exp$ function shows $$\int_0^\infty e^{-ax}e^{c x^b } dx =\int_0^\infty e^{-ax}\sum_{n=0}^{\infty} \dfrac{c^n x^{nb}}{n!} dx = \sum_{n=0}^{\infty} \dfrac{c^n\Gamma(1+nb)}{a^{1+nb}n!}$$

Answer 2

$$\int_0^\infty e^{cx^b}e^{-ax}\ \mathrm{dx}$$ $$=\int_0^\infty \sum_{k=0}^\infty \frac{c^kx^{bk}}{k!}e^{-ax}\ \mathrm{dx}$$ $$=\sum_{k=0}^\infty \frac{c^k}{k!}\int_0^\infty x^{bk}e^{-ax}\ \mathrm{dx}$$ $$=\sum_{k=0}^\infty \frac{c^k\Gamma(bk+1)}{k!a^{bk+1}}$$ $$=\frac1a\sum_{k=0}^\infty \frac{\Gamma(bk+1)}{k!}\left(\frac{c}{a^b}\right)^k$$ $$=\frac1a\ _1\hspace{-2px}\Psi_0\left[\begin{array}{c|}(1, b)\\-- \end{array}\ \frac{c}{a^{b}}\right]$$ Where $\Psi$ is the Fox-Wright function.